Probability of printing without a skip. The old printer in the computer room skips, on average, 1 character in 25 (i.e., the
chance of skipping any particular character is 1/25).
First part :
Determine the probability that a line of 45 characters is printed without a skip.
for the first part I thought that the probability of printing without a skip is 1-1/25, for the 45 characters : (24/25)^45 but it confuses me, I think I need to solve it with geometric series but not sure
Second part:
Determine the probability that at least one character is skipped in the first 12.
For the second part 1- probability (without a skip in the first 12)
1- (24/25)^12
Really unsure about this problem, can someone help me?
 A: Since the problem only gives the average probability, you are supposed to
assume the Poisson distribution according to which
\begin{equation*}
P(x)=\frac{e^{-\lambda }\lambda ^{x}}{x!}\text{, where }\lambda \text{ is
the average number of events.}
\end{equation*}
Determine the probability that a line of 45 characters is printed without a skip: Since $1$ error occurs on average for $25$
characters,
\begin{equation*}
\lambda =45/25=1.8\text{ skips per }45\text{ characters on average.}
\end{equation*}
Then,
\begin{equation*}
P(x=0)=\frac{e^{-1.8}\times (1.8)^{0}}{0!}=e^{-1.8}\approx 0.16530\text{.}
\end{equation*}
Determine the probability that at least one character is
skipped in the first $12$: This time
\begin{equation*}
\lambda =12/25=0.48\text{ skips per }12\text{ characters on average.}
\end{equation*}
Note that,
\begin{equation*}
P(x=0)=\frac{e^{-0.48}\times (1.8)^{0}}{0!}=e^{-0.48}
\end{equation*}
and we have
\begin{equation*}
P(x\geq 1)=1-P(x=0)=1-e^{-0.48}\approx 0.38122\text{.}
\end{equation*}
All the Best.
